# How to calculate screening constant for a given element ????

Screening constant can be found by Slater’s rules which are as follows:

1.The electronic configuration of the given element is written as follows :

(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) (5d)…

2. Electrons to the right of the electron of interest (ns and np group) do not contribute to shielding constant

3. All other electrons in the ns and np group contribute to an extent of 0.35 each to the screening constant except 1s group which contributes to an extent of 0.30.

4. All the electrons in (n-1) th shell contribute 0.85 each to the screening constant

5. All the electrons in (n-2) th or lower shell contribute 1.00 each to the screening constant

For d and f electrons rule 1 to 3 are same.In place of 4 and 5 rule 6 is applicable

6. All the electrons in the groups lying left to the nd and nf group contribute 1.0 each to the screening constant

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1. Write down the electronic configuration of the element as shown below.

• (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) (5d)…

• Fill the electrons according to Aufbau principle.
• Any electrons to the right of the electron of interest do not contribute to shielding constant.

• The shielding constant for each group is formed as the sum of the following contributions:
• All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units except 1s group, where the other electron contributes only 0.30.

• If the group is of the [s, p] type, an amount of 0.85 from each electron (n-1) shell and an amount of 1.00 for each electron from (n-2) and lower shell.

• If the group is of the [d] or [f] type then an amount of 1.00 for each electron from all lying left to that orbital.

2. 2
For example: (a) Calculate effective nuclear charge in Nitrogen for 2p electron.
• Electronic configuration- (1s2) (2s2, 2p3).

• Screening constant, σ = (0.35 × 4) + (0.85 × 2) = 3.10

• Effective nuclear charge, Z* = Z – σ = 7 – 3.10 = 3.90

3. 3
(b) Calculate effective nuclear charge and screening constant seen in 3p electron in Silicon.
• Electronic configuration- (1s2) (2s2, 2p6)(3s2, 3p2).

• σ = (0.35 × 3) + (0.85 × 8) + (1 × 2) = 9.85

• Z* = Z – σ = 14 – 9.85 = 4.15

4. 4
(c) Calculate effective nuclear charge in Zinc for 4s electron & for 3d electron.
• Electronic configuration- (1s2) (2s2, 2p6)(3s2, 3p6)(3d10)(4s2).

• For 4s electron,

• σ = (0.35 × 1) + (0.85 × 18) + (1 × 10) = 25.65

• Z* = Z – σ = 30 – 25.65 = 4.35

• For 3d electron,

• σ = (0.35 × 9) + (1 × 18) = 21.15

• Z* = Z – σ = 30 – 21.15 = 8.85

5. 5
(d) Calculate effective nuclear charge on one of 6s electron in tungsten. (At. No. =74)
• Electronic configuration- (1s2) (2s2, 2p6)(3s2, 3p6)(4s2, 4p6) (3d10) (4f14) (5s2, 5p6)(5d4), (6s2)

• σ = (0.35 × 1) + (0.85 × 12) + (1 × 60) = 70.55

• Z* = Z – σ = 74 – 70.55 =3.45

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The difference between the atomic number of an element and the apparent atomic number for a given process; this difference results for screening

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